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IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P0. Choose the correct option(s) from the following:

A
Internal energies at A and B are the same.
B
Work done by the gas in process AB is P0V0 ln4.
C
Pressure at C is P0/4.
D
Temperature at C is T0/4.

Explanation

The internal energy of one mole of an ideal gas at temperature T is given by U = 3RT/2. The process AB is isothermal i.e., TA = TB = T0, which makes UA = UB.

The work done in the isothermal process AB is

WAB = nRT ln(VB/VA)

= nRT0 ln(4V0/V0) = p0V0 ln4.

Information regarding p and T at C can not be obtained from the given graph. Unless it is mentioned that line BC passes through origin or not.

Hence, the correct options are (a) and (b).

Note:

If we assume that the line BC pass through the origin. In the process BC, the slope V/T is constant. Thus, BC is an isobaric process (as V/T = nR/p, by the ideal gas equation). Thus,

pC = pB = RTB / VB

= RT0/(4V0) = (RT0/V0)4 = p0/4.

Apply the ideal gas equation for the states A and C to get

$${T_C} = {{{p_C}{V_C}} \over {{p_0}{V_0}}}{T_0} = {{({p_0}/4){V_0}} \over {{p_0}{V_0}}}{T_0} = {T_0}/4$$.

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